Use the definition limit definition of a derivative:
If thederivative is 0 then \[f \prime(2)=\lim_{h \rightarrow 0}=\frac{ f(2+h)-f(2) }{ h }\]
If this is zero than This implies that f(2+h)-f(2)=0, so thinking about the defintion of the slope:
\[m=\frac{ f(2+h)-f(2) }{ h }=\frac{ 0 }{ h }=0\]

you could prove this using the definition of the derivative:\[f'(x)=\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\]Another way to say this is:\[f'(x)=\lim_{\Delta x \rightarrow 0}\frac{\Delta y}{\Delta x}=\frac{dy}{dx}\]Graphically, this is the slope of the tangent line at the point as @Sujay said.|dw:1354847377868:dw|